Here's the question you clicked on:
Dodo1
Find the equation of the tangent line to the curve (a lemniscate) 2(x^2+y^2)2=25(x^2−y^2) at the point (−3,1). The equation of this tangent line can be written in the form y=mx+b where m is:
First Differentiate the function to find the slope.
thank you but i dont know process. i am not sure about calculation, particulary differentiation
I mean i know the process
You know the process, but you need assistance with the differentiation portion?
Yes, when it comes to both chain rule and products. I get really confused
\[\large 2(x^2+y^2)^2=25(x^2−y^2)\]So we'll be differentiating this `implicitly`, meaning, `as is`. No solving for y ahead of time. When we execute the chain rule, a few y' will pop out here and there, and we'll want to solve for that variable. \(\large y'(-3,1)=m\). This is the m in our equation \(\large y=mx+b\). Hmm so let's see...
It appears we have no product rule which is nice.
We should be able to get a numerical value for m, no x's or y's in our m. I'm not really sure what that Y7 is.
Taking the derivative of the left side gives us, `taking the derivative of the outermost function first, the square on the brackets`:\[\large 2\cdot2(x^2+y^2)^1\color{royalblue}{\frac{d}{dx}(x^2+y^2)}\]The blue is due to the chain rule. Understand that part?
\[\large 2\cdot2(x^2+y^2)\color{royalblue}{(2x+2yy')}\]So we get something like this on the left side.
i lied, I know product rule, power rule etc, but i dont know what the chain rule is
mmmm i always have trouble explaining that one. It's very unintuitive.
could you give me example of chain rule problem maybe?
The definition says this,\[\large \frac{d}{dx}f\left(g(x)\right) \qquad = \qquad f'(g(x))\cdot g'(x)\]When you take the derivative of this, a function within a function, you ~Differentiate the outer function, f in this case. ~Then you make a copy of the inner function, and multiply by it's derivative. Here is a silly example. \[\large \frac{d}{dx}\left(x^2\right)^{4}\] In this example we're thinking of \(\large g(x)=x^2\). And \(\large f(x)=x^4\). If you remember back to doing compositions of functions, you would understand that when we plug g(x) INTO f(x), our f(x), \(\large f(\color{royalblue}{x})=\left(\color{royalblue}{x}\right)^4\) will become, \(\large f(\color{royalblue}{g(x)})=\left(\color{royalblue}{g(x)}\right)^4\) which is \(\large f(\color{royalblue}{x^2})=\left(\color{royalblue}{x^2}\right)^4\). Applying the chain rule gives us,\[\large f'(g(x))=4\left(g(x)\right)^3 g'(x)\]See how the rule tells us to make a copy of the inner function and take it's derivative?\[\large f'(g(x))=4\left(x^2\right)^3 (x^2)'\]\[\large f'(g(x))=4\left(x^2\right)^3 (2x)\]
Oo.. I see, Thank you. It was very clear! really cool :)
If you were to simplify this down, you would get,\[\large f'(g(x))=8x^7\] Which we know is correct, because we could have easily approached this example without using the chain rule.\[\large y=(x^2)^4=x^8\]And we know the derivative will match what we got from the chain rule example.
It's a weird concept :d you're like.. making copies of stuff.. but teachers never use that type of language to describe it. So I dunno lol
no, you explain well. i always get confused and could not tell wether i should use chain or product.
So we applied the derivative of the left side,\[\large \frac{d}{dx}(thing)^2 \qquad \rightarrow \qquad 2(thing)(thing)'\] right? :O
haha thats even more clear haha def, taking note
could we back to the first question?
This one? :o \[\large 2(x^2+y^2)^2=25(x^2−y^2)\]
I see the chain rule now, (x^2+y^2)^2?
\[\large 2(x^2+y^2)\]That's the derivative of the outer function yes? The square on the brackets. But the chain rule is telling us to make a copy of the inside, and take it's derivative.\[\large 2(x^2+y^2)(x^2+y^2)'\]
\[\large 2(x^2+y^2)(2x+2y)\]So we should get somethingggggggg like this, yes?
:O why two in the fron of the bracket? I understand the (2x+2y) tho
You applied the power rule to the outer function \(\large \left(\qquad \right)^2\)
power rule and chain double? :O
Power rule the outside, make a copy of the inside, take derivative of copy.
Since we took our derivative with respect to x, any time we differentiate a variable that IS NOT x, like y for example, we have to apply the chain rule, and multiply by y'.
\[\large 2(x^2+y^2)(2x+2y\color{royalblue}{y'})\]See this blue term? We took the derivative of a variable that is not x, namely y, so a y' popped out.
mmmmmm.... Ok. I think :/
so whenever its chain rule i put y' in the end?
No. Let's not think of that portion as chain rule, I think that's just going to confuse you. Whenever we take the derivative of y, we will multiply by y'. We don't throw it on the end, we always attach it to the y that we're taking the derivative of.
right, Ok. i will remember that by my heart.
Try taking the derivative of the right side, tell me what you get.
Ignore the 25 in front. No chain rule, just take the derivative of the stuff in the brackets.
close, but you took the derivative of a y, you need to multiply that part by y'
its still funny how i have to put y' in the end. is it for implicit derviation?
yes. If we had an equation like this,\[\large y=x^2\]The derivative gives us,\[\large y'=2x\]Ok that makes sense.. The y' is telling us what the derivative of y is. But what if we had something like this?\[\large y^2=x^2\]Taking the derivative,\[\large 2y=2x\] Hmm we don't have a derivative term do we?
so 2(x^2+y~2)(2x+2y)=25(2x-2yy')
Wasn't there a 2 in front of the equation on the left at the start?? Or was that just a typo?
2(x^2+y^2)^2=25(x^2−y^2)
See that 2 at the front? That would give us an extra factor of 2 in our derivative.
the question equation is 2(x^2+y^2)^2=25(x^2-y^2)
\[4(x^2+y^2)(2x+2y)=25(2x-2yy')\]
\[\large 4(x^2+y^2)(2x+2y\color{royalblue}{y'})=25(2x-2yy')\]Woops you missed a y'.
ops! haha thats derivative of the first equation?
yes. We took the derivative of y on each side, so each y will get a y'
Hmmm from here, I think it's going to be easier to FIRST plug in your coordinate \(\large (-3,1\;)\) and THEN solve for y'.
plug both number into the equation?
yes, -3 for all the x's. 1 in for all the y's.
and don't plug anything in for y'. That's what we'll be solving for.